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zhannawk [14.2K]
3 years ago
13

3. one fifth as a decimal

Mathematics
1 answer:
Y_Kistochka [10]3 years ago
6 0
The answer for this question is 0.2
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Stacy rolls a pair of six-sided fair dice.
insens350 [35]

Answer:

Pr(the sum of the numbers rolled is either a multiple of 3 or an even number)=\frac{2}{3}

Step-by-step explanation:

Let A be the event "sum of numbers is multiple of 3"

and B be the event "sum is an even number".

As our dice has six sides, so the sample space of two dices will be of 36 ordered pairs.

|sample space | = 36

Out of which 11 pairs have the sum multiple of 3 and 18 pairs having sum even.

So Pr(A)= \frac{11}{36}

and Pr(B)= \frac{18}{36}

and Pr(A∩B) = \frac{5}{36}, as 5 pairs are common between A and B.

So now Pr(A or B)= Pr(A∪B)

                            = Pr(A)+Pr(B) - Pr(A∩B)

                            = \frac{11}{36} + \frac{18}{36} - \frac{5}{36}

                            = \frac{24}{36}

                            = \frac{2}{3}

7 0
3 years ago
First to solve right !!!!WIL GET MARKED BRAINLIEST!!!! (easy)
fredd [130]

Answer:

183$

Step-by-step explanation:

.........................

3 0
2 years ago
Read 2 more answers
How do you do this question?
Lena [83]

Step-by-step explanation:

The Taylor series expansion is:

Tₙ(x) = ∑ f⁽ⁿ⁾(a) (x − a)ⁿ / n!

f(x) = 1/x, a = 4, and n = 3.

First, find the derivatives.

f⁽⁰⁾(4) = 1/4

f⁽¹⁾(4) = -1/(4)² = -1/16

f⁽²⁾(4) = 2/(4)³ = 1/32

f⁽³⁾(4) = -6/(4)⁴ = -3/128

Therefore:

T₃(x) = 1/4 (x − 4)⁰ / 0! − 1/16 (x − 4)¹ / 1! + 1/32 (x − 4)² / 2! − 3/128 (x − 4)³ / 3!

T₃(x) = 1/4 − 1/16 (x − 4) + 1/64 (x − 4)² − 1/256 (x − 4)³

f(x) = 1/x has a vertical asymptote at x=0 and a horizontal asymptote at y=0.  So we can eliminate the top left option.  That leaves the other three options, where f(x) is the blue line.

Now we have to determine which green line is T₃(x).  The simplest way is to notice that f(x) and T₃(x) intersect at x=4 (which makes sense, since T₃(x) is the Taylor series centered at x=4).

The bottom right graph is the only correct option.

3 0
2 years ago
Only I need help with 1,2 and three please
Mazyrski [523]
Is that the whole thing?

4 0
2 years ago
Can anyone solve this ASAP WORTH 100 points.
yKpoI14uk [10]

Answer:

yea i can help

Step-by-step explanation:

7 0
2 years ago
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